8. | Torque | = (0.125 m)*F =
I*|alpha|, assuming a force of magnitude F is applied at the rim
and is tangent to wheel.
Alpha can be found from: 0 = (220)*6.28*/60 rad/s - |alpha
|*time, where time = 45.0 seconds. Find |alpha| and note alpha is
negative. Substitute into the first equation to get |torque| and
also F , the magnitude of the slowing friction force. Consult table 9.2
to get I.

11. See example 10.3 and class notes.

12.You could use energy methods to solve :
Go to Quiz 9 and see a string of
problems using it in different ways. You pick the best way for this
problem; the correct choice is clear. See:
42, 43, 49, 50, 52, 51, Quiz 9
. In particular see 42. USE CONSERVATION OF ENERGY :
(1/2)*mvi2 + (1/2)*I*wi2
+ mgyi
= (1/2)*m*vf2 + (1/2)*I*wf2
+ mgyf. , where I = (1/2)*M*R2, and I
is the unknown moment of inertia of the
cylindrical pulley. Note: w =v/R. Radius R and final linear speed
of the hanging mass are all given and m = 15.0 kg. The system begins
from rest so initial kinetic energies are zero but please
note yi = 4.00 m and final value of y is ZERO.
The final linear speed is 3.50 m/s.
(a) Find final w = v/R after falling distance 4.00 m from rest.
(b) Find I.

42. See class notes. I did this in
class !! See also example 10.13. Note in that example the cable leans
rightward instead of leftward as in figure 10.60. See also this sample
exam problem. After clicking the link scroll down to
problem 5. Problem 42 is just like the sample exam problems except
you must add the torque exerted by the weight of the load at the
right end of the beam. That torque takes the form L*Mg, where M is
the load mass. Here you are given Mg =
300 N. Thus , 0 = L*T*sin 37 - (L/2)*mg - L*Mg , where m is the mass of
the beam. Solve for tension T.

43. Check back soon but try
finding examples that might look like this last turn in, including
likenesses to previous Quiz 10 problems.